A test contains 3 sections each containing 4 questions. A student need

Question

A test contains 3 sections each containing 4 questions. A student needs to solve a total of 8 questions, ensuring he/she picks at least 2 questions from each section. If a student follows the rules of the test, what is the probability that he/she will pick all 4 questions of a section?
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A. 3/11
B. 3/10
C. 3/7
D. 1/2
E. 3/4

gmatophobia · Answer

The test comprises of three sections and each section has four questions.

Let's name these sections  S_1, S_2  and  S_3 . It is given that each section has 4 questions.

We need to select 2 questions from each section, hence the following two selection patterns are possible

Pattern 1  : 4 - 2 - 2

Select four questions from any one section (out of the three available sections) and select two questions from the remaining two sections.

Note: This is set of favorable cases

Pattern 2  : 3 - 3 - 2

Select three questions from any two sections (out of the three available sections) and select two questions from the remaining section.

Calculating the number of ways the selection can be made

Pattern 1  : 4 - 2 - 2

Step 1  : Out of the three available sections  S_1, S_2  and  S_3  identify one section from which 4 questions can be selected.

This can be done in  ^3C_1   = 3 ways

Step 2  : From the section selected in Step 1, select 4 four questions from that section

This can be done in  ^4C_4   = 1 way

Step 3  : From the remaining two sections, select two questions

This can be done in  ^4C_2 * ^4C_2   = 6 * 6 ways

Sub Total = 3 * 6 * 6

Pattern 2  : 3 - 3 - 2

Step 1  : Out of the three available sections  S_1, S_2  and  S_3  identify two sections from which 3 questions can be selected.

This can be done in  ^3C_2   = 3 ways

Step 2  : From the sections selected in Step 1, select 3 four questions from each section

This can be done in  ^4C_3 * ^4C_3   = 4 * 4 ways

Step 3  : From the remaining section, select two questions

This can be done in  ^4C_2   = 6 ways

Sub Total = 3 * 4 * 4 * 6

Total Cases = (3 * 6 * 6)+ (3 * 4 * 4 * 6)

Required Probability = \frac{ 	ext{Favorable Cases} }{ 	ext{Total Cases}}

=  \frac{(3 * 6 * 6) }{ (3 * 6 * 6)+ (3 * 4 * 4 * 6)}

Simplifying we get

P =  \frac{3}{11}

Option A

BNDN · Answer

Why isn't the total no.of cases possible 12C8?

Bunuel · Answer

Read the question carefully:

A test contains 3 sections each containing 4 questions. A student needs to solve a total of 8 questions,   ensuring he/she picks at least 2   questions from each section. If a student follows the rules of the test, what is the probability that he/she will pick all 4 questions of a section? 
 
The total can’t be 12C8, because the condition “at least 2 from each section” restricts the selection.

12C8 counts  all  ways to pick 8 questions out of 12, including cases where one section contributes 0 or 1 question. Those cases violate the rule.

That’s why the solution splits into patterns (4-2-2 and 3-3-2), which are the only distributions that satisfy “at least 2 per section.”

A test contains 3 sections each containing 4 questions. A student need

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